Answers

2014-08-15T19:07:19+05:30
Suppose D(x) = {1 if rational, 0 if irrational} 

If the limit of a function as it approaches some value a is defined and equals L, then for any epsilon>0 there exists a delta>0 such that if 0<|x-a|<delta then |f(x)-L)|<epsilon. 

i) choose some rational a. Then we want to show that the limit as D(x) approaches a is 1. Suppose the limit exists. Let epsilon=0.5. Based on the limit definition, there should be some delta so that all x within delta of a have D(x) values within 0.5 of 1. However, no matter which epsilon we choose, i.e. no matter how small of an interval we choose around a, we can find an irrational number inside that interval because the irrational numbers are dense in the real numbers. This is a contradiction, because the D(x) value of that irrational number is 0, which is more than 0.5 away from 1. 

i) the proof works similarly if you choose an irrational number, since the rationals are dense in the reals.
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